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A path along the edges of a graph that traverses every vertex exactly once and terminates at its starting point. Also known as Hamiltonian circuit; Hamiltonian cycle.

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What is the significance of the Hamiltonian path problem in graph theory and its applications in various fields?

The Hamiltonian path problem in graph theory is significant because it involves finding a path that visits each vertex exactly once in a graph. This problem has applications in various fields such as computer science, logistics, and network design. It helps in optimizing routes, planning circuits, and analyzing connectivity in networks.


How can the Hamiltonian path be reduced to a Hamiltonian cycle?

To reduce a Hamiltonian path to a Hamiltonian cycle, you need to connect the endpoints of the path to create a closed loop. This ensures that every vertex is visited exactly once, forming a cycle.


What is the significance of a Hamiltonian cycle in a bipartite graph and how does it impact the overall structure and connectivity of the graph?

A Hamiltonian cycle in a bipartite graph is a cycle that visits every vertex exactly once and ends at the starting vertex. It is significant because it provides a way to traverse the entire graph efficiently. Having a Hamiltonian cycle in a bipartite graph ensures that the graph is well-connected and has a strong structure, as it indicates that there is a path that visits every vertex without repeating any. This enhances the overall connectivity and accessibility of the graph, making it easier to analyze and navigate.


How can the Hamiltonian cycle be reduced to a Hamiltonian path?

To reduce a Hamiltonian cycle to a Hamiltonian path, you can remove one edge from the cycle. This creates a path that visits every vertex exactly once, but does not form a closed loop like a cycle.


What is the difference between NP and NP Complete problems?

- a problem in NP means that it can be solved in polynomial time with a non-deterministic turing machine - a problem that is NP-hard means that all problems in NP are "easier" than this problem - a problem that is NP-complete means that it is in NP and it is NP-hard example - Hamiltonian path in a graph: The problem is: given a graph as input, an algorithm must say whether there is a hamiltonian path in it or not. in NP: here is an algorithm that works in polynomial time on a non-deterministic turing machine: guess a path in the graph. Check that it is really a hamiltonian path. NP-hard: we use reduction from a problem that is NP-comlete (SAT for example). Given an input for the other problem we construct a graph for the hamiltonian-path problem. The graph should have a path iff the original problem should return "true". Therefore, if there is an algorithm that executes in polynomial time, we solve all the problems in NP in polynomial time.j


How does the concept of a vertex cover relate to the existence of a Hamiltonian cycle in a graph?

In graph theory, a vertex cover is a set of vertices that covers all edges in a graph. The concept of a vertex cover is related to the existence of a Hamiltonian cycle in a graph because if a graph has a Hamiltonian cycle, then its vertex cover must include at least two vertices from each edge in the cycle. This is because a Hamiltonian cycle visits each vertex exactly once, so the vertices in the cycle must be covered by the vertex cover. Conversely, if a graph has a vertex cover that includes at least two vertices from each edge, it may indicate the potential existence of a Hamiltonian cycle in the graph.


Is there a way to demonstrate that the problem of determining whether a given path exists in a graph is NP-complete?

Yes, the problem of determining whether a given path exists in a graph can be demonstrated as NP-complete by reducing it to a known NP-complete problem, such as the Hamiltonian path problem. This reduction shows that the path existence problem is at least as hard as the known NP-complete problem, making it NP-complete as well.


How can the 3-SAT problem be reduced to the Hamiltonian cycle problem in polynomial time?

The 3-SAT problem can be reduced to the Hamiltonian cycle problem in polynomial time by representing each clause in the 3-SAT problem as a vertex in the Hamiltonian cycle graph, and connecting the vertices based on the relationships between the clauses. This reduction allows for solving the 3-SAT problem by finding a Hamiltonian cycle in the constructed graph.


Can a graph have an Euler circuit but not a Hamiltonian circuit?

Yes. An example: _____A---------B________ A connected directly to B and D by one path. _____|_______/|\________ B connected directly to A and E by one path, and to C by two paths. _____|______/_|_\_______ _____|_____/___\_|______ _____|__E/_____\|______ E connected directly to B and D by one path. _____|____\_____C______ C connected directly to B and D by two paths. _____|_____\____|\_____ _____|______\___|__\___ _____|_______\__|__/___ _____|________\_|_/____ _____|_________\|/_____ _____-------------D_____ D connected directly to A and E by one path, and to C by two paths. There is an Euler circuit: ABCDEBCDA But a Hamiltonian circuit is impossible: as part of a circuit A can only be reached by the path BAD, but once BAD has been traversed it is impossible to get to both C and E without returning to B or D first. However there is a Hamiltonian Path: ABCDE.


What touches every vertex exacaly once but does not return to its origin?

Hamiltonian path


Can a graph have a euler circuit but not a hamiltonian circuit?

Yes. Example: .................................................... ...A * ........................................... ......|.\ ......................................... eg Euler circuit: ACDCBA ......|...\ ........... --------- ............. ......|.....\........./...............\............ The Hamilton circuit is impossible as it has two ......|.......\...../...................\.......... halves (ACD & CD) connected to each other only ......|.........\./.......................\........ at vertex C. Once vertex C has been reached in ......|.......C *........................* D.... one half, it can only be used to start a path in ......|........./.\......................./......... the other half, or complete the cycle in the ......|......./.....\.................../........... current half; or if the path starts at C, it will end ......|...../.........\.............../............. without the other half being visited before C is ......|.../ ........... --------- .............. revisited. ......|./ ........................................... ...B *.............................................. ......................................................


What is the shortest path in an undirected graph?

The shortest path in an undirected graph is the path between two vertices that has the smallest total sum of edge weights.